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3 years ago

What is this question?? (needing help)

Mathematics

2 answers:

Free_Kalibri [48]3 years ago

5 0

468 x 0.001 = 0.468

46.8 x 0.1 = 4.68

4.68 x 10^3 = 4680

0.468 x 10^2 = 46.8

Hope this helps!

Sav [38]3 years ago

4 0

0.468 is the right answer on the first part. First you multiply without the decimal point, and then put the decimal point as the answer. And it gave us, $468*1=468$ , or therefore, the answer has three decimal places. And it gave us the answer is 0.46768 is the right answer. 4.68 is the right answer on the second part.  You can also multiply without the decimal point, and then put the decimal point it's going to be the answer. And it gave us, $468*1=468$ , or therefore, the answer it has two decimal places and the answer is 4.68 is the right answer. 4680 is the right answer on the third part. First you had to multiply by the exponents, and it gave us, $10^3=10*10*10=1000*4.68$ . Multiply by the numbers, and it gave us the answer is $4.68*1000=4680$ , or 4680 is the right answer. 46.8 is the right answer on the last part. First you had to multiply by the exponents, and it gave us, $10^2=10*10=100*0.468$ . Then you multiply by the numbers, and it gave us, $0.468*100=46.8$ , and it gave us the answer is 46.8 is the right answer. Hope this helps! And thank you for posting your question at here on brainly, and have a great day. -Charlie

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3 years ago

A ship leaves port at 10 miles per hour, with a heading of N 35° W. There is a warning buoy located 5 miles directly north of th

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The value of the angle subtended by the distance of the buoy from the

port is given by sine and cosine rule.

The bearing of the buoy from the is approximately 307.35°

Reasons:

Location from which the ship sails = Port

The speed of the ship = 10 mph

Direction of the ship = N35°W

Location of the warning buoy = 5 miles north of the port

Required: The bearing of the warning buoy from the ship after 7.5 hours.

Solution:

The distance travelled by the ship = 7.5 hours × 10 mph = 75 miles

By cosine rule, we have;

a² = b² + c² - 2·b·c·cos(A)

Where;

a = The distance between the ship and the buoy

b = The distance between the ship and the port = 75 miles

c = The distance between the buoy and the port = 5 miles

Angle ∠A = The angle between the ship and the buoy = The bearing of the ship = 35°

Which gives;

a = √(75² + 5² - 2 × 75 × 5 × cos(35°))

By sine rule, we have;

$\displaystyle \frac{a}{sin(A)} = \mathbf{ \frac{b}{sin(B)}}$

Therefore;

$\displaystyle sin(B)= \frac{b \cdot sin(A)}{a}$

Which gives;

$\displaystyle sin(B) = \mathbf{\frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }}$

$\displaystyle B = arcsin\left( \frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }\right) \approx 37.32^{\circ}$

Similarly, we can get;

$\displaystyle B = arcsin\left( \frac{75 \cdot sin(35^{\circ})}{\sqrt{75^2 + 5^2 - 2 \times 75\times5\times cos(35^{\circ}) } }\right) \approx \mathbf{ 142.68^{\circ}}$

The angle subtended by the distance of the buoy from the port, C is therefore;

C ≈ 180° - 142.68° - 35° ≈ 2.32°

By alternate interior angles, we have;

The bearing of the warning buoy as seen from the ship is therefore;

Bearing of buoy ≈ 270° + 35° + 2.32° ≈ 307.35°

Learn more about bearing in mathematics here:

brainly.com/question/23427938

5 0

2 years ago